Surprising Results (and respond please)
9th October 2006
This past Friday, my school had an inservice day for the teachers. I was responsible for a small portion of the day. It was my task to provide a demonstration of the Four Corners differentiated instruction strategy to the rest of the faculty. I decided the best way to accomplish that was to actually go through the activity instead of just telling my colleagues about it. So of course, I based the activity on a mathematical statement. Usually, any discussion of mathematics with the faculty in general is met by moans, groans, and eye-rolling. This time, I was surprised. But I’ll get to that in a minute.Four Corners works like this:
- Teacher writes a debateable or controversial statement on the board.
- Students move to the corner of the room where the sign is posted that most closely represents their opinion on the statement.
- Discussion, questions, and debate are then allowed between the four groups.
- Next, students are allowed to switch to another corner if they changed their opinion.
- Students are invited to explain what caused any change of opinion.
The statement I posed to the faculty was this:
There are more integers total (… -3, -2, -1, 0, 1, 2, 3, …) than there are decimal numbers just between 0 and 1.
The four statements that I posted in the four corners were:
- I agree with this statement
- No, the opposite is true
- The two sets of numbers are equal in number
- It cannot be determined
I was quite surprised by the amount of discussion that my statement generated. It was very interesting to see how “non-math” people think about mathematics in a purely theoretical setting. The faculty really “dug” the discussion and many of them switched corners as thoughts and arguements were shared.
Ultimately, they ended up fairly evenly divided amongst the four corners. By the end of the activity, they were begging me to tell them which response was actually the correct one (I had told them that one and only one of the responses was indeed correct). I eventually did tell them. But in this post, I’ll leave it to one (or more) of my readers to leave a comment as to which response should be chosen to the above statement.
October 11th, 2006 at 10:00 am
The answer is always C. infinity = infinity
?
October 11th, 2006 at 12:31 pm
Depends on what you mean by “more” for infinite sets. Assuming you mean cardinality, the answer is (b).
October 11th, 2006 at 12:59 pm
Yes, Xian, I did mean “cardinality” by “more”. Thanks for the correct response. Do you care to elaborate?
Sorry Ben, it’s my fault. I accidentally put the answers in the wrong order. I didn’t mean to mess you up. Of course the right answer is always C. I need to remember that when making tests for my students!
October 11th, 2006 at 1:59 pm
I should, of course, stipulate that I like answer (c) more. And that this proof is in no way a demonstration of my brilliance, just a proof I’ve seen a dozen times:
Consider an ordering of Z: 0, 1, -1, 2, -2, 3, -3, … will do. Now I can index the elements in the integers; I can talk about the “first” or the “fifteenth” integer and so on.
Let’s say I want to identify a subset of the integers. I’d like to do it by considering an infinite sequence; that is, a list of numbers a_1, a_2, … established by this rule: if the i’th integer (by the ordering I gave above) is included in this particular subset, then a_i is equal to 1. Otherwise, a_i is equal to 0.
I do this for each subset of the integers (ignoring the fact that I can’t, you know, ever stop). Now, I contend that this set of sequences is “bigger,” or has higher cardinality, than the integers.
I prove it this way: say that they have the same cardinality, or there are as many integers as there are infinite sequences that I designed above. Then I can make a list of the sequences, much as I made a list of the integers before. For each integer, I go through and assign a single sequence to it.
Now I’m going to invent a new sequence (b_1, b_2, …). The rule for my new sequence is this: in the list of integers, if the sequence associated with the i’th integer has a_i = 0, then I let b_i = 1. I let b_i = 0 otherwise.
Now, we see that my sequence (b_1, b_2, …) is not accounted for in any of the previous sequences, for it differs from the i’th sequence at the i’th place. But that means that we didn’t get all the possible sequences when we made our association above. And since each sequence corresponds to a subset of the integers, we are missing one of those! That means that there can’t be as many integers as there are subsets of the integers.
Well, who cares about that? You asked about the real line. Well, consider all of the numbers between 0 and 1 in base 2. Each of them can be written as an infinite sequence 0.(c_1 c_2 c_3 …) of zeroes and ones. So for any number between 0 and 1, we can put it into correspondence with the number of sequences containing just 0s and 1s, and there are “more” of those sequences than there are integers, so there are more reals between 0 and 1 than there are integers.
October 29th, 2006 at 10:19 pm
I am not sure I understand “decimal numbers just between 0 and 1.”
Are you including terminating decimals only, or are you including non-terminating decimals?
October 30th, 2006 at 8:21 am
“decimal numbers just between 0 and 1″ meant “real numbers between 0 and 1.” I just wanted to state it in approximate lay terms at the time.